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FODE_1-10

# FODE_1-10 - Tutorial FODE Ordinary First-Order Differential...

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6. Solutions to Exercises Exercise 1. The antiderivative of 2 x is x 2 : Thus, the general solution is y ( x ) = x 2 + C: The condition tells us 0 =1 2 + C = ) C = ¡ 1 ; or y ( x ) = x 2 ¡ 1 : Exercise 2. Writing ln A insteadof C as the constant of integration, we can write Eqn. 4 as ln j y ln A = ln μ j y j A = x 2 2 = ) y = A exp μ x 2 2 : Exercise 3. y (0)= 2 = ) 2 = A ¢ 1 = ) A = 2 : Thus, y ( x )= 2exp μ x 2 2 : Exercise 4. You will find examples in virtually every chapter, but you might want to go first to the sections on Newton’s 2nd Law, motion in viscous media, the charging of capacitors and inductors and radioactive decay. Exercise 5. Eqn. 10 separates immediately as dy y = ¡ q ( x ) dx: Exercise 6. a. Nonlinear, because of the y 2 factor in the second term. The equation is separable: dy y 2 = ¡ 2 xdx; with solution y ( x ) = 1 x 2 + C : b. Linear, inhomogeneous and separable. dy

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FODE_1-10 - Tutorial FODE Ordinary First-Order Differential...

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